The potential energy u of a particle varies with distance x from a fixed origin as u=A x^1/2/x+B where A&B are constants. what will be the dimensions of AB
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Potential energy ,
[math]U=((A)x^{1/2})/(x+B)[/
math]
We know dimension for energy
[U]=
[[math]ML^2T^{-2}[/math]]
For two quantities to be added
the have to be the same
dimension. Hence B has
dimensions of x.
=> [B]=L
Hence [U]=[math][[/math]
[math]ML^2T^{-2}]=(([A])L^
{1/2}/L)[/math]
[math]=[A]L^{-1/2}[/math]
=>
[math][A]=ML^{5/2}T^{-2}[/
math]
Note: [X] refers to the dimension
of X.
[math]U=((A)x^{1/2})/(x+B)[/
math]
We know dimension for energy
[U]=
[[math]ML^2T^{-2}[/math]]
For two quantities to be added
the have to be the same
dimension. Hence B has
dimensions of x.
=> [B]=L
Hence [U]=[math][[/math]
[math]ML^2T^{-2}]=(([A])L^
{1/2}/L)[/math]
[math]=[A]L^{-1/2}[/math]
=>
[math][A]=ML^{5/2}T^{-2}[/
math]
Note: [X] refers to the dimension
of X.
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